# Math Help - Need help in squared differential EQn.

1. ## Need help in squared differential EQn.

So my problem is this: "A non-negative function f has the property that its ordinate set over an arbitrary interval has an area proportional to the arc length of the graph above the interval. Find f"

So i started with the integral of f and the definition of arc length in 2-space integral ( sqrt(1 + f ' (x)^2)) and then set them equal to each other. but then i get this:

f(x) ^ 2 = 1 + f ' (x) ^ 2

But I don't know how to solve this. Any help will be great.

2. Originally Posted by hashshashin715
So my problem is this: "A non-negative function f has the property that its ordinate set over an arbitrary interval has an area proportional to the arc length of the graph above the interval. Find f"

So i started with the integral of f and the definition of arc length in 2-space integral ( sqrt(1 + f ' (x)^2)) and then set them equal to each other. but then i get this:

f(x) ^ 2 = 1 + f ' (x) ^ 2

But I don't know how to solve this. Any help will be great.
You can do one of two things. First, solve for $f'$ so

$f' = \pm \sqrt{f^2-1}$ which is now separable.

Second, differentiate your ODE giving

$2 f f' = 2 f' f'' \; \text{or}\; f'(f''-f) = 0$ which gives two cases $f' = 0$ or $f''-f=0$.

Although I need to ask, where is your constant of proportionality?

3. Thanks a lot for the help. And I thought the constants were suppose to some afterward in the general form of the solution.

4. Originally Posted by hashshashin715
Thanks a lot for the help. And I thought the constants were suppose to some afterward in the general form of the solution.
You do but the question talks about proportionality so you'll need a constant for that, i.e.

$f = k \sqrt{1 + f'^2}$